The general notion at work here is the completion of a measure.
$\newcommand{\R}{\mathbb{R}} \newcommand{\B}{\mathcal{B}}$
Let's write $\B$ for the Borel $\sigma$-algebra on $\R$. If $\mu$ is a positive Borel measure on $\R$ (i.e. a countably additive set function $\mu : \B \to [0,\infty]$), let $\B_\mu$ be the $\sigma$-algebra generated by $\B$ together with the sets $\{A \subset B : B \in \B, \mu(B) = 0\}$ (i.e. throw in all subsets of sets with measure zero). This is called "completing $\B$ with respect to $\mu$", and of course $\mu$ has a natural extension to $\B_\mu$. When we take $\mu$ to be Lebesgue measure $m$, $\B_m$ is precisely the Lebesgue $\sigma$-algebra.
In this notation, I think your questions are as follows:
Is $\B_m \subset \B_\mu$ for every $\mu$?
If not, is there a measure $\mu$ with $\B_\mu = \B$?
For 1, the answer is no. As you suspect, the Cantor measure $\mu_C$ is a counterexample. If $C$ is the Cantor set and $f : C\to [0,1]$ is the Cantor function, then we can write $\mu_C(B) = m(f(B))$. If $A \notin \B_m$ is a non-Lebesgue measurable set, then $f^{-1}(A) \notin \B_{\mu_C}$. But $f^{-1}(A) \subset C$ and $m(C) = 0$, so $f^{-1}(A) \in \B_m$.
For the second question, the answer is yes, sort of. One example is counting measure $\mu$ which assigns measure 1 to every point (hence measure $\infty$ to every infinite set). Here the only set of measure 0 is the empty set, which is already in $\B$, so $\B_\mu = \B$. Another example is a measure which assigns measure 0 to every countable (i.e. finite or countably infinite) set, and measure $\infty$ to every uncountable set. Now the measure zero sets are all countable, hence so are all their subsets, but all countable sets are already Borel.
Note these are not Lebesgue-Stieltjes measures, because they give infinite measure to every nontrivial interval.
In fact, suppose $\mu$ is a measure such that $\B_\mu = \B$. Then I claim every uncountable Borel set $B$ has $\mu(B) = \infty$. Suppose $B$ is an uncountable Borel set. It is known that such $B$ must have a subset $A$ which is not Borel. If $\mu(B) = 0$, then $A \in \B_\mu \backslash \B$, which we want to avoid. So we have to have $\mu(B) > 0$. On the other hand, it is also known that an uncountable Borel set can be written as an uncountable disjoint union of uncountable Borel sets. Each of these must have nonzero measure, so this forces $\mu(B) = \infty$. In particular $\mu$ is not Lebesgue-Stieltjes.
So in fact, any Lebesgue-Stieltjes measure has measure-zero sets with non-Borel subsets, and hence can be properly extended by taking the completion.
A closely related idea is that of universally measurable sets, which are those sets $B$ which are in $\B_\mu$ for every finite (or, equivalently, every $\sigma$-finite) measure $\mu$. There do exist universally measurable sets which are not Borel. On the other hand, the above example with Cantor measure shows that there are Lebesgue measurable sets which are not universally measurable.
In addition to the previous advice, note that the function you gave does not "approximate" $f$. An approximating sequence $\{s_n\}$ for $f$ (which $f$ would have iff $f$ were measurable provided the measure space for the domain of $f$ is complete) would need to be within a distance of $\epsilon$ from $f$ for any given $\epsilon >0$. However, the function $s$ you gave as an "approximation" cannot approximate $f$, in the sense that the sequence $\{s_n\}$ with $s_n=s$ $ \forall n \in \mathbb{N}$ gets no closer than $1$ from $\chi_{[0,1]}$.
Again, it is important to note that a function $f$ on a complete measure space $X$ is measurable if and only if $f$ is the pointwise limit of some sequence of simple functions -or, trivially, is a simple function itself. Consequently, any sequence of "simple" functions approximating a nonmeasurable function must contain a "simple" function with the characteristic function of a nonmeasurable set as part of its construction. In a sense, this is why you must include the assumption that $f$ is measurable in your definition for the Lebesgue integral.
To wit, recall that the integral of a characteristic function is the measure of the pullback set in your domain; in the case of your $f$, since $f$ is characteristic, the integral, were it to be defined, is the measure of the pullback, $$\int f d\mu = \mu\{f^{-1}\{1\}\}=\mu\{[0,1]\},$$
but you have not defined the measure for $[0,1]$, which is not even in your $\sigma$-algebra; nor can we infer the measure of $[0,1]$ from the definition you gave for your measure space, as your collection of measurable subsets is already a closed $\sigma$-algebra that is $\sigma$-finite under $\mu$ (hence, your measure space cannot even be extended, in the usual Caratheodory way, to include $[0,1]$ with an accompanying well-defined measurement).
As a curiosity tangential to your question, it is possible for nonmeasurable functions to arise from limits of simple functions in a complete measure space, but such a collection of simple functions must be uncountable. For instance, with Lebesgue measure on $\mathbb{R}$, take $f=\chi_V$ to be the characteristic function of the (uncountable and nonmeasurable) Vitali set $V$ on $[0,1]$, and consider the (uncountable) collection of measurable functions $\{\chi_v\}, v\in V$. Then $\chi_V=\sup \{\chi_v\}_{v\in V}$. Were we to define an integral as you wish, then in this case, you may want to say that the integral $\int \chi_V = \sup \{ \int \chi_v \} = 0$. But, again, since $\chi_V$ is itself characteristic, we should then have $\mu(V)=0$, but $\mu(V)$ is not defined for $V$ under the Lebesgue measure.
To address your request for a resource, see Royden's Real Analysis, 4th ed., chapters 17 and 18 (particularly pp. 362-363 were helpful as a reference to me for this post).
Best Answer
The only $\mathcal{S}$-measurable functions (assuming that we take the Borel $\sigma$-algebra on the codomain) are the constant functions, thus $f \colon x \mapsto \chi_{[0,1]}(x)\cdot x$ isn't measurable.
Hence if the definition of the Lebesgue integral for non-negative functions demands the measurability, $\int f\,d\lambda$ is not defined.
Some authors define $$\int g \,d\mu = \sup \: \biggl\{ \int s\, d\mu : 0 \leqslant s \leqslant g, s \text{ simple}\biggr\}$$ even for non-measurable functions $g \geqslant 0$. With that definition we would have $\int f\,d\lambda = 0$ since $0$ is the largest simple measurable functions $\leqslant f$.